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How did it became 10pi/27 from the equation encircled?

How did it became 10pi/27 from the equation encircled?

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter12: Quadratic Functions
Section12.6: Solving Problems Involving Quadratic Equations
Problem 7P
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How did it became 10pi/27 from the equation encircled?
Express your answer in rectangular form both in EXACT value and approximate value. Round off approximate values to 5 decimal places. For each number, graph the roots in an Argand diagram showing the magnitude and angle of each of the roots. So there will be 2 Argand diagrams, one of no.1, the other separate diagram for no.2. 1. Find the nine roots of -256 – j256/3 .
Transcribed Image Text:Express your answer in rectangular form both in EXACT value and approximate value. Round off approximate values to 5 decimal places. For each number, graph the roots in an Argand diagram showing the magnitude and angle of each of the roots. So there will be 2 Argand diagrams, one of no.1, the other separate diagram for no.2. 1. Find the nine roots of -256 – j256/3 .
+2-T-2 2 +2-T:2 •k = 2: V52 (c 2 (cos () + i sin () = 2 cos COS + i sin 9 9. 10T 10T 1UT 10T + 2i sin 27 27 27 27
Transcribed Image Text:+2-T-2 2 +2-T:2 •k = 2: V52 (c 2 (cos () + i sin () = 2 cos COS + i sin 9 9. 10T 10T 1UT 10T + 2i sin 27 27 27 27
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